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SubjectRe: [PATCH] sched: Improve readability in update_cpu_load() code
2008/5/15 Srivatsa Vaddagiri <vatsa@linux.vnet.ibm.com>:
> On Thu, May 15, 2008 at 06:34:59PM +0530, Gautham R Shenoy wrote:
>> Author: Gautham R Shenoy <ego@in.ibm.com>
>> Date: Thu May 15 17:55:49 2008 +0530
>>
>> sched: Improve readability in update_cpu_load() code
>>
>> Currently the cpu_load[i] is calculated as:
>> this_rq->cpu_load[i] = (old_load*(scale-1) + new_load) >> i;
>>
>> However, since scale = 2^i, this can be simplified as:
>> this_rq->cpu_load[i] = old_load + ((new_load - old_load) >> i);
>>
>> Makes it easier to read.
>> Signed-off-by: Gautham R Shenoy <ego@in.ibm.com>
>>
>> diff --git a/kernel/sched.c b/kernel/sched.c
>> index 2d7d8f1..e1a6985 100644
>> --- a/kernel/sched.c
>> +++ b/kernel/sched.c
>> @@ -2921,7 +2921,7 @@ static void update_cpu_load(struct rq *this_rq)
>> */
>> if (new_load > old_load)
>> new_load += scale-1;
>> - this_rq->cpu_load[i] = (old_load*(scale-1) + new_load) >> i;
>> + this_rq->cpu_load[i] = old_load + ((new_load - old_load) >> i);
>
> This wont work when new_load < old_load ..
>
> For ex: I tried this prog:
>
> #include <stdio.h>
>
> main()
> {
> unsigned long old_load = 100, new_load = 90, this_load, this_load1;
> int i = 1, scale = 2 << i;
>
> this_load = (old_load*(scale-1) + new_load) >> i;
> this_load1 = old_load + ((new_load - old_load) >> i);

it should be

this_load2 = 2 * old_load + (new_load >> i) - (old_load >> i);

scale == 2 << i == 1 << (i + 1), so scale >> i = 2.

and it may result in (probably) little difefrences due to different rounding.
I think, for me the currently existing version is more easy-to-read.


> --
> Regards,
> vatsa

--
Best regards,
Dmitry Adamushko


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